Re: [閒聊] 每日leetcode

作者: DJYOSHITAKA (Evans)   2024-05-30 20:54:08
差不多
找subarray xor是0的 然後++
直接2D forloop嚕過去
應該可以用prefix-xor省
懶改了
def countTriplets(self, arr: List[int]) -> int:
cnt = 0
n = len(arr)
for i in range(n-1):
cur_xor = arr[i]
for j in range(i+1,n):
cur_xor = cur_xor^arr[j]
if cur_xor == 0:
cnt += (j-i)
return cnt
作者: Che31128 (justjoke)   2024-05-30 20:58:00
還在卷

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